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I'm working on the following exercise:

"Let $\Sigma:=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2=z^2+1,-1\leq z\leq3\}$ be a surface with orientation $\omega$.

Compute $\int\int_{\Sigma}(\nabla\times F)\cdot\omega\ dS$ knowing that $F=(-\frac{y}{3},-\frac{z}{3},-\frac{x}{3})$ and $\omega (0,1,0)=(0,1,0)$. Verify the result you have obtained using Stokes' Theorem."

Here's what I've done:

-> $\nabla\times F=(\frac{1}{3},\frac{1}{3},\frac{1}{3})$;

-> I've divided the surface in three pieces:

(+) $0

$\omega_+=\varphi_{+_x}\times\varphi_{+_y}=(-\frac{x}{\sqrt{x^2+y^2-1}},-\frac{y}{\sqrt{x^2+y^2-1}},1)$

($0$)$\ z=0$: $\omega_0=(x,y,0)$ (ok, since $\omega (0,1,0)=(0,1,0)$;

(-)$\varphi_{-}:=(x,y,-\sqrt{x^2+y^2-1})\to \varphi_{-_x}=(1,0,-\frac{x}{\sqrt{(x^2+y^2-1)}}), \varphi_{-_y}=(0,1,-\frac{y}{\sqrt{(x^2+y^2-1)}})\to$

$\omega_-=\varphi_{-_x}\times\varphi_{-_y}=(\frac{x}{\sqrt{x^2+y^2-1}},\frac{y}{\sqrt{x^2+y^2-1}},1)$

(Notice that I've not normalized the normal vector since when computing the flux of a vector field across a surface in space we have $\int\int_S F\cdot n dS=\int\int_S F\cdot (\varphi_x \times\varphi_y)dxdy$)

So $\int\int_{\Sigma_+}(\nabla\times F)\cdot\omega_+ dS=\int\int_{\Sigma_+}(\frac{1}{3},\frac{1}{3},\frac{1}{3})\cdot (-\frac{x}{\sqrt{x^2+y^2-1}},-\frac{y}{\sqrt{x^2+y^2-1}},1)dxdy=\frac{1}{3}\int\int_{\Sigma_+}(-\frac{x}{\sqrt{x^2+y^2-1}}-\frac{y}{\sqrt{x^2+y^2-1}}+1)=?$

Here is where I get stuck, since I can't figure out what the bounds for this double integral should be; can anyone give me an hint about this?

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By the divergence theorem, your integral equals $$ \Phi = \iiint_E \underbrace{\nabla\cdot \nabla \times F}_{=0}\; dV-\iint_{S_1}\nabla \times F\; dS-\iint_{S_2}\nabla \times F\; dS, $$ where $S_1$ and $S_2$ are the surfaces that close $\Sigma$ at $z=-1$ and $z=3$. $S_1$ can be parametrized as follows: $$ \begin{cases} x=x\\ y=y\quad \mbox{with }\; x^2+y^2\le 2\\ z=-1 \end{cases} $$ And similarly for $S_2$: $$ \begin{cases} x=x\\ y=y\quad \mbox{with }\; x^2+y^2\le 10\\ z=3 \end{cases} $$ It follows that $$ \Phi = -\iint_{x^2+y^2\le 2} \pmatrix{1/3\\1/3\\1/3}\cdot \pmatrix{0\\0\\-1}\; dS -\iint_{x^2+y^2\le 10} \pmatrix{1/3\\1/3\\1/3}\cdot \pmatrix{0\\0\\1}\; dS =\frac{A(S_1)}{3}-\frac{A(S_2)}{3} = -\frac{8\pi}{3} $$

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    Thank you for your answer. To check the answer with Stokes' theorem I have used as $C$ the circumference at $z=-1$ i.e. $x^2+y^2=2$ so I get $\oint_{x^2+y^2=2} F\cdot dr =\int_{0}^{2\pi} (-\frac{\sqrt{2}\sin (t)}{3}, 0,-\frac{\sqrt{2}\cos (t)}{3})\cdot (-\sqrt{2}\sin (t),\sqrt{2}\cos (t),0)dt=\frac{2}{3}\pi$ which should be equal to $\int\int_{\Sigma} (\nabla \times F)\cdot dS + \int\int_{\Sigma (2)}(\nabla\times F)\cdot dS$ so $\int\int_{\Sigma} (\nabla\times F)\cdot dS=\frac{2}{3}\pi - \frac{A(S_2)}{3}=-\frac{8}{3}\pi$; can you tell me why I'm getting the wrong sign? – 2017-01-21
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    I subtracted the $\int\int_{\Sigma (2)}$ part from $\oint$ because I don't know if Stokes' theorem applies only to the boundaries of closed surfaces; is that the case or can it be applied to any boundary of any surface? – 2017-01-21
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    I think it is just an orientation problem. $\Sigma_2$ should be orientated downwards, so it should give you $-\frac{-A(S_2)}{3}$ and $C$ should be orientated counterclockwise. – 2017-01-21
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    Can you also tell me how did you choose the normal vectors for $S_1$ and $S_2$? I would have said $(0,0,1)$ for $S_2$ and $(0,0,-1)$ for $S_1$ but you did the opposite, why? – 2017-01-21
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    Woops, you are correct, that is the right orientation. This yields $-8\pi/3$ which matches your answer. I have edited accordingly. – 2017-01-21